Consider the polar curve $r=\tan(\theta)$. What is the slope of the tangent line to the curve $r$ when $\theta = \dfrac{\pi}{4}$ ? Give an exact expression. $\text{slope }=$
Solution: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\tan(\theta)}\cos(\theta) \\\\ y&={\tan(\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{\sec^2(\theta)\sin(\theta)+\sin(\theta)}{\cos(\theta)} \end{aligned}$ Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{4}}$. $\begin{aligned} \left. \dfrac{dy}{dx}\right| _{\theta =\tfrac{\pi }{4}}&=\dfrac{\sec^2\left({\dfrac{\pi}{4}}\right)\sin\left({\dfrac{\pi}{4}}\right)+\sin\left({\dfrac{\pi}{4}}\right)}{\cos\left({\dfrac{\pi}{4}}\right)} \\\\ &=\dfrac{(\sqrt{2})^2\left(\dfrac{\sqrt{2}}{2}\right)+\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}} \\\\ &=\dfrac{2\left(\dfrac{\sqrt{2}}{2}\right)}{\dfrac{\sqrt{2}}{2}}+\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}} \\\\ &=2+1 \\\\ &=3 \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{4}$ equals $3$. The graph of the tangent is shown.